2–14 RPN: The Automatic Memory Stack
File name 33s-E-Manual-1008-Publication(1st).doc Page : 386
Printed Date : 2003/10/8 Size : 13.7 x 21.2 cm
The above example, when solved left–to–right, needed all registers in the stack at
one point:
Keys: Display: Description:
4
Ï
14
Ï
Saves 4 and 14 as intermediate
numbers in the stack.
7
Ï
3
_
At this point the stack is full with
numbers for this calculation.
¸
Intermediate result.
Ù
Intermediate result.
2
Ã
Intermediate result.
¯
Final result.
More Exercises
Practice using RPN by working through the following problems:
Calculate:
(14 + 12)
×
(18 – 12)
÷
(9 – 7) = 78.0000
A Solution:
14
Ï
12
Ù
18
Ï
12
Ã
¸
9
Ï
7
Ã
¯
Calculate:
23
2
– (13
×
9) + 1/7 = 412.1429
A Solution:
23
=
13
Ï
9
¸
Ã
7
,
Ù
Calculate:
5961.0)7.05.12()8.04.5(
3
=−÷×
Solution:
5.4
Ï
.8
¸
.7
Ï
3
)
12.5
w
Ã
¯ ?
or
5.4
Ï
.8
¸
12.5
Ï
.7
Ï
3
)
Ã
¯
?
Calculate:
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